Show that the function y defined implicitly by xy =

Show that the function $y$ defined implicitly by $xy = \ln(ey)$ satisfies the equation $y^{2}+(xy-1)y'=0$.

Step 1: Simplify the right-hand side

The logarithm of a product splits, and $\ln e = 1$: $\ln(ey) = \ln e + \ln y = 1 + \ln y$.

So the relation is

\[ xy = 1 + \ln y. \tag{1} \]

Step 2: Differentiate with respect to $x$

The left side needs the product rule; the right side needs the chain rule: $y + xy' = \frac{y'}{y}$.

Step 3: Clear the denominator

Multiply through by $y$ (legitimate, since $\ln y$ requires $y>0$): $y^{2} + xyy' = y'$.

Step 4: Move everything to one side

\[ y^{2} + xyy' - y' = 0 \;\Longrightarrow\; y^{2} + \bigl(xy-1\bigr)y' = 0. \]

This is exactly the required equation, so the proof is complete.

Step 5: Verification

Solving (1) for $x$ gives an explicit parametrisation of the curve by $y$: \[x = \frac{1+\ln y}{y} \;\Longrightarrow\; \frac{dx}{dy} = \frac{\dfrac1y\cdot y-(1+\ln y)}{y^{2}} = \frac{-\ln y}{y^{2}}\]

\[y' = \frac{1}{dx/dy} = -\frac{y^{2}}{\ln y} = \frac{y^{2}}{1-xy} \quad\text{since } \ln y = xy-1 \;\;\checkmark\]

\[y = e:\ x = \frac{2}{e} \approx 0.73576,\quad y' = -\frac{e^{2}}{1} \approx -7.38906; \qquad y = e^{-1}:\ x = 0,\quad y' = e^{-2} \approx 0.13534\]